If the sides $a, b, c $ form a successive geometric progression with common ratio $r(r>1)$, then prove that $A<B<\frac{\pi}{3}<C$

Question

From the given data, the obvious result that can be drawn is

$$b^2=ac$$

Also $$\cos B =\frac{a^2+c^2-b^2}{2ac}$$ $$\cos B =\frac{a^2+c^2}{2b^2}-\frac{b^2}{2b^2}$$ $$\cos B =\frac{(a+c)^2}{2b^2}-1-\frac 12$$

I don’t know how to proceed.

Answer 1

You are so closeee

$$ \begin{aligned} \frac{a^{2}+c^{2}-b^{2}}{2ac}&=\frac{a^{2}+c^{2}-ac}{2ac}\\ &>\frac{2ac-ac}{2ac} \end{aligned} $$

Thus $\angle A<\angle B < \frac{\pi}{3}$

Answer 2

According to the sine rule, the geometric progression implies

$$\sin^2 B = \sin A \sin C = \frac12( \cos(A-C) + \cos B )$$

which leads to

$$2\cos^2 B + \cos B -2 = -\cos(A-C)>-1$$

or

$$(2\cos B -1)(\cos B +1) > 0\implies \cos B > \frac12$$

which yields $B < \frac\pi3$. Thus, $A<B<\frac{\pi}{3}<C$.